这个PHP代码有什么问题？ 从数据库检索数据

Question

What is wrong in this code? Here I am trying to show the data fetched from a mySQL database when someone clicks on the "Show" button (Which is a submit button).

<?php
if(isset($_POST['submit'])){
    $mysqli = new mysqli("localhost", "rms", "sarangi", "rms");
    if($mysqli === false){
        die("Connection Error.");
    }
    else{
        echo'<table border="1"><tr><th> Name </th><th> Username </th></tr>';
        $sql = "SELECT * FROM users";
        if($result = $mysqli->query($sql)){
            if($result->num_rows > 0){
                while($data = $result->fetch_array()){
                    echo'<tr><td>' . $data['firstname'] . " " .     $data['lastname'] . '</td><td>' . $data['username'] . '</td></tr>';
                }
            }
        }
        echo'</table>';
    }
    $mysqli->close();
}
else{
?>
<form action="index.php" method="POST">
   <input type="submit" id="submit" value="Show" />
</form>
<?php
  }
?>

Answer 1

You have to name your input field

<input type="submit" id="submit" value="Show" />

should be

<input name="submit" type="submit" id="submit" value="Show" />

Answer 2

change your code to this

<form action="" method="POST">
   <input type="submit" id="submit" name="submit" value="Show" />
</form>

I have changed form action to "" since its going to submit to the same page. or you can leave it as index.php if this page is index.php. You need to put the name attribute of input submit as submit